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QUASI-EXACT DIFFERENTIAL EQUATION
By Benny Le Van-Vietnam
Abstract
This article examines a specific variant of exact differential equation whose form is a matrix
product of exact partial derivatives and a modifying vector.
1. Introduction
( ) ( )
It is widely perceived (see e.g. [1]) that if ܲ ݔ,ݕ and ܳ ݔ,ݕ are two bivariable function
such that:
[ ( )] [ ( )]
߲ ܲ ݔ,ݕ =߲ܳ ݔ,ݕ (1)
߲ݕ ߲ݔ
Then the exact first-ordered ordinary differential equation:
( ) ( ) (2)
ܲ ݔ,ݕ ݀ݔ+ܳ ݔ,ݕ ݀ݕ = 0
Has solutions as:
ܷ(ݔ,ݕ) = ܿ݊ݏݐ
⎧ ௫ ௬
⎪
⎡ ( ) ( ) ( )
⎪ܷ ݔ,ݕ = නܲ ݔ,ݕ ݀ݔ+ නܳ ݔ,ݕ ݀ݕ+ܿ݊ݏݐ
⎢ (3)
⎢ ௫ ௬
బ బ
⎨ ௬
⎢ ௫
⎢
⎪ܷ(ݔ,ݕ)= නܲ(ݔ,ݕ)݀ݔ+ නܳ(ݔ ,ݕ)݀ݕ+ܿ݊ݏݐ
⎪
⎢
⎩⎣ ௫ ௬
( ) బ బ
Of which, ܷ ݔ,ݕ is the potential function of (1) where exactness of the above differential
equation is determined by the criteria that:
[ ( )] [ ( )]
߲ ܷ ݔ,ݕ ߲ ܷ ݔ,ݕ
⎧ =ܲ(ݔ,ݕ) =ܳ(ݔ,ݕ)
⎪ ߲ݔ ߲ݕ (4)
ଶ[ ( )] [ ( )] [ ( )]
⎨ ߲ ܷ ݔ,ݕ =߲ ܲ ݔ,ݕ =߲ ܳ ݔ,ݕ
⎪
⎩ ߲ݔ߲ݕ ߲ݕ ߲ݔ
Accordingly, we define the quasi-exact differential equation (QDE) as:
( ) ( ) ( ) (5)
ܲ ݔ,ݕ ݀ݔ+ܴ ݔ,ݕ ܳ ݔ,ݕ ݀ݕ = 0
[ ( ) ( ) ]
An alternate form of QDE is ݒݓ = 0 where ݒ = ܲ ݔ,ݕ ݀ݔ ܳ ݔ,ݕ ݀ݕ are exact partial
( ) ʹ
[1 ܴ ݔ,ݕ ] ( )
derivatives, ݓ = is the modifying vector, and ܴ ݔ,ݕ is the modifier.
( )
We shall solve (5) where ܴ ݔ,ݕ is (i) a constant in Section 2; and a variable function in
Section 3, and Section 4, respectively. Following, Section 5 provides discussion.
1 RMM-QUASI-EXACT DIFFERENTIAL EQUATIONS-BENNY LE VAN
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2. Constant quasi-exact differential equation
( ) 1
In the case ܴ ݔ,ݕ is a constant ݎ (ݎ ≠ 0 and ݎ ≠ 1) , the constant QDE becomes:
ܲ(ݔ,ݕ)݀ݔ+ݎܳ(ݔ,ݕ)݀ݕ = 0 (6)
We shall transform (6) to the exact form by multiplying both sides with an integrating factor
( )
ܵ ݔ,ݕ which is not a constant. Equation (6) becomes:
( ) ( ) ( ) ( ) (7)
ܲ ݔ,ݕ ܵ ݔ,ݕ ݀ݔ +ݎܳ ݔ,ݕ ܵ ݔ,ݕ ݀ݕ = 0
( )
It is supposed to find ܵ ݔ,ݕ such that:
[ ( ) ( )] [ ( ) ( )]
߲ ܲ ݔ,ݕ ܵ ݔ,ݕ =߲ݎܳ ݔ,ݕ ܵ ݔ,ݕ
߲ݕ ߲ݔ
⇔߲ܵܲ+߲ܲܵ=ݎ൬߲ܵܳ+߲ܳܵ൰
߲ݕ ߲ݕ ߲ݔ ߲ݔ
߲ܲ ߲ܳ ߲ܵ ߲ܵ
⇔ܵ൬ −ݎ ൰+ܲ −ݎܳ =0
߲ݕ ߲ݔ ߲ݕ ߲ݔ
( )
With ܷ ݔ,ݕ as determined under (3) and (4), we could rewrite the above as:
߲ଶܷ ߲ଶܷ ߲ܷ߲ܵ ߲ܷ߲ܵ
ܵቆ −ݎ ቇ+ −ݎ =0
߲ݔ߲ݕ ߲ݔ߲ݕ ߲ݔ߲ݕ ߲ݔ߲ݕ
( ) ߲ଶܷ ( )߲ܷ߲ܵ
⇔ܵ1−ݎ + 1−ݎ
߲ݔ߲ݕ ߲ݔ߲ݕ
⇔߲ܵଶܷ+߲ܷ߲ܵ=0
( ) ଶ
Since ܵ ݔ,ݕ ≠ ܿ݊ݏݐ, we could divide both sides of the above by ߲ܵ and consequently
obtain a second-ordered ordinary differential equation:
߲ଶܷ ߲ܷ ݀ଶܷ ܷ݀ (8)
ܵ + =0⇔ܵ + =0
߲ܵଶ ߲ܵ ݀ܵଶ ݀ܵ
In equation (8), replacing ܶ = ܷ݀⁄݀ܵ gives:
݀ܶ ݀ܶ ݀ܵ | | | | ݇
ܵ +ܶ=0⇔ =− ⇔lnܶ =−lnܵ +ܿ⇔ܶ=
݀ܵ ܶ ܵ ܵ
Henceforth, we obtain:
ܷ݀ ݇ ݀ܵ
| |
= ⇔ܷ݀=݇ ⇔ܷ=݇lnܵ +݈⇔ܵ=ܽ݁
݀ܵ ܵ ܵ
The finding that ܵ = ܽ݁ gives solutions of (7) are ܹ(ݔ,ݕ) = ܿ݊ݏݐ, of which:
௫ ௬
( ) ( ) ( ) ( ) ( )
ܹ ݔ,ݕ = නܲ ݔ,ݕ ܵ ݔ,ݕ ݀ݔ+ݎ නܳ ݔ,ݕ ܵ ݔ,ݕ ݀ݕ+ܿ݊ݏݐ
௫ ௬ (9)
బ బ
௫ ( ) ௬ ( )
(௫,௬ ) ߲ܷ ݔ,ݕ (௫,௬) ߲ܷ ݔ,ݕ
= නܽ݁ బ ݀ݔ+ݎ නܽ݁ ݀ݕ+ܿ݊ݏݐ
߲ݔ ௬ୀ௬ ߲ݕ
బ
௫ ௬
Or: బ బ
1 If ݎ = 0, solutions are ܲ(ݔ,ݕ) = 0 or ݔ = ܿ݊ݏݐ; if ݎ = 1, the QDE becomes an exact differential equation.
2 RMM-QUASI-EXACT DIFFERENTIAL EQUATIONS-BENNY LE VAN
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௫ ௬
ܹ(ݔ,ݕ)= නܲ(ݔ,ݕ)ܵ(ݔ,ݕ)݀ݔ+ݎ නܳ(ݔ ,ݕ)ܵ(ݔ ,ݕ)݀ݕ+ܿ݊ݏݐ
௫ ௬ (10)
బ బ
௫ ( ) ௬ ( )
(௫,௬) ߲ܷ ݔ,ݕ (௫ ,௬) ߲ܷ ݔ,ݕ
= නܽ݁ ݀ݔ+ݎ නܽ݁ బ ݀ݕ+ܿ݊ݏݐ
߲ݔ ߲ݕ
௫ୀ௫
௫ ௬ బ
బ బ
Example 1
Solve the following differential equation:
(ݕ݁௫ +݁௬)݀ݔ = (ݔ݁௬ +݁௫)݀ݕ (11)
Equation (11) is quasi-exact where ܲ(ݔ,ݕ) = ݕ݁௫ +݁௬, ܳ(ݔ,ݕ) = ݔ݁௬ + ݁௫, and ݎ = −1.
( ) ௬ ௫
Besides, formula (3) gives ܷ ݔ,ݕ = ݔ݁ +ݕ݁ .
It is found that the integrating factor is:
ܵ(ݔ,ݕ) = ܽ݁(௫,௬) = ܽ݁(௫ା௬ೣ)
Thus, solution of Example 1 is ܹ(ݔ,ݕ) = ܿ݊ݏݐ, where:
௫ ௬
(௫బା௬ ೣ) ௫ ௬ (௫ା௬ೣ) ௬ ௫
( ) బ ( బ) ( )
ܹݔ,ݕ = නܽ݁ ݕ ݁ +݁ ݀ݔ− නܽ݁ ݔ݁ +݁ ݀ݕ+ܿ݊ݏݐ
௫ ௬
బ బ
ܽ ( బ ೣ) ܽ ( ೣ) ௬
௫ ା௬ ௫ ௫ ା௬
( ) బ
ܹݔ,ݕ = ݁ |௫ − ݁ | +ܿ݊ݏݐ
బ ௬
ܾ ܾ బ
2ܽ ೣ ܽ ೣ
(௫ బା௬ ) (௫ ା௬ )
( ) బ
ܹݔ,ݕ = ݁ − ݁ +ܿ݊ݏݐ
ܾ ܾ ( )
Simplifying ܽ = ܾ, then a solution of (11) is ܹ ݔ,ݕ = ܿ݊ݏݐ, where:
(௫బା௬ ೣ) (௫ା௬ೣ)
ܹ(ݔ,ݕ) = 2݁ బ −݁ +ܿ݊ݏݐ
An alternate expression is:
ೣ ೣ
ܹ(ݔ,ݕ) = ݁(௫ ା௬ ) −2݁(௫బ ା௬ బ) +ܿ݊ݏݐ
3. Univariable quasi-exact differential equation
We shall solve the QDE (5) when ܴ(ݔ,ݕ) ≠ ܿ݊ݏݐ. A unique case is:
( )
߲ ܴܳ =߲ܳ⇔߲ܴܳ+ܴ߲ܳ=߲ܳ⇔߲ܴܳ=(1−ܴ)߲ܳ⇔ ߲ܴ =߲ܳ
߲ݔ ߲ݔ ߲ݔ ߲ݔ ߲ݔ ߲ݔ ߲ݔ 1−ܴ ܳ
The above case results in a separable differential equation:
ܴ݀ ݀ܳ | | | |
1−ܴ= ܳ ⇔ln1−ܴ =lnܳ +ܿ⇔1−ܴ=݇ܳ⇔ܴ=1−݇ܳ
Of which, ݇ = ܿ݊ݏݐ. In this case, (5) becomes exact whose solution is under the form of (3).
For ܴ(ݔ,ݕ) ≠ 1−݇ܳ(ݔ,ݕ), it is supposed to find the integrating factor ܵ(ݔ,ݕ) ≠ ܿ݊ݏݐ
such that:
( ) ( ) ( ) ( ) ( )
ܲ ݔ,ݕ ܵ ݔ,ݕ ݀ݔ +ܴ ݔ,ݕ ܳ ݔ,ݕ ܵ ݔ,ݕ ݀ݕ = 0
[ ( ) ( )] [ ( ) ( ) ( )]
ቐ ߲ ܲ ݔ,ݕ ܵ ݔ,ݕ =߲ܴ ݔ,ݕ ܳ ݔ,ݕ ܵ ݔ,ݕ (12)
߲ݕ ߲ݔ
Finding ܵ(ݔ,ݕ):
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߲ܵܲ+߲ܲܵ=߲ܴܳܵ+ܴ߲ܵܳ+ܴ߲ܳܵ
߲ݕ ߲ݕ ߲ݔ ߲ݔ ߲ݔ
߲ଶܷ ߲ܷ߲ܵ ߲ଶܷ ߲ܷ߲ܵ ߲ܴ߲ܷ (13)
⇔ܵ + =ܴܵ +ܴ +ܵ
߲ݔ߲ݕ ߲ݔ߲ݕ ߲ݔ߲ݕ ߲ݔ߲ݕ ߲ݔ߲ݕ
⁄ ( ) ( ) ⁄ ( )
This includes a specific case that ߲ܴ ߲ݔ = 0 ⇔ ܴ ݔ,ݕ = ܴ ݕ or ߲ܴ ߲ݔ = 0 ⇔ ܴ ݔ,ݕ =
( ) ( )
ܴ ݔ . In this case, the process of finding ܵ ݔ,ݕ turns equivalent to Section 2.
( )
Henceforth, the QDE (5) is comprehensively solvable if ܴ ݔ,ݕ is a univariable function.
( ) ( )
Without loss of generality, we assume ܴ ݔ,ݕ = ܴ ݕ and (13) becomes:
߲ଶܷ ߲ܷ߲ܵ ߲ଶܷ ߲ܷ߲ܵ
ܵ + =ܴܵ +ܴ
߲ݔ߲ݕ ߲ݔ߲ݕ ߲ݔ߲ݕ ߲ݔ߲ݕ
⇔߲ܵଶܷ+߲ܷ߲ܵ=ܴ߲ܵଶܷ+ܴ߲ܷ߲ܵ
⇔ܵ(1−ܴ)߲ଶܷ+(1−ܴ)߲ܷ߲ܵ=0
Since ܴ ≠ ܿ݊ݏݐ and ܵ ≠ ܿ݊ݏݐ, we could transform the above equation as:
߲ଶܷ ߲ܷ ݀ଶܷ ܷ݀
| |
ܵ + =0⇔ܵ + =0⇔ܷ=݇lnܵ +݈⇔ܵ=ܽ݁
߲ܵଶ ߲ܵ ݀ܵଶ ݀ܵ
( )
Thus, solutions of (5) are ܹ ݔ,ݕ = ܿ݊ݏݐ, where:
௫ ௬
( ) ( ) ( ) ( ) ( ) ( )
ܹ ݔ,ݕ = නܲ ݔ,ݕ ܵ ݔ,ݕ ݀ݔ+ නܴ ݕ ܳ ݔ,ݕ ܵ ݔ,ݕ ݀ݕ+ܿ݊ݏݐ
௫ ௬ (14)
బ బ
௫ ߲ܷ(ݔ,ݕ) ௬ ߲ܷ(ݔ,ݕ)
(௫,௬ ) (௫,௬)
= නܽ݁ బ ݀ݔ+ නܴܽ(ݕ)݁ ݀ݕ+ܿ݊ݏݐ
߲ݔ ௬ୀ௬ ߲ݕ
బ
௫ ௬
Or: బ బ
௫ ௬
ܹ(ݔ,ݕ)= නܲ(ݔ,ݕ)ܵ(ݔ,ݕ)݀ݔ+ නܴ(ݕ)ܳ(ݔ ,ݕ)ܵ(ݔ ,ݕ)݀ݕ+ܿ݊ݏݐ
௫ ௬ (15)
బ బ
௫ ߲ܷ(ݔ,ݕ) ௬ ߲ܷ(ݔ,ݕ)
(௫,௬) ( ) (௫బ,௬)
= නܽ݁ ݀ݔ+ නܴܽ ݕ ݁ ݀ݕ+ܿ݊ݏݐ
߲ݔ ߲ݕ
௫ ௬ ௫ୀ௫బ
బ బ
Example 2
Solve the QDE:
ݕ ( ) (16)
ቀlnݕ+ ቁ݀ݔ+ ݔ+ݕlnݔ ݀ݕ=0
ݔ
Equation (16) is quasi-exact where:
ܲ(ݔ,ݕ) = lnݕ +ݕ
⎧
⎪ ݔ
ܳ(ݔ,ݕ) = lnݔ +ݔ
⎨ ݕ
⎪ ܴ(ݔ,ݕ) = ݕ
⎩
The exact differential formula gives:
4 RMM-QUASI-EXACT DIFFERENTIAL EQUATIONS-BENNY LE VAN
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